Question: Graph this system of equations and solve. $y = x + 2$ $7x-3y = 6$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Solution: The y-intercept for the first equation is $2$ , so the first line must pass through the point $(0, 2)$ The slope for the first equation is $1$ . Remember that the slope tells you rise over run. So in this case for every $1$ position you move up You must also move $1$ position to the right. $1$ position to the right. $1$ position up from $(0, 2)$ is $(1, 3)$ Graph the blue line so it passes through $(0, 2)$ and $(1, 3)$ Convert the second equation, $7x-3y = 6$ , to slope-intercept form. $y = \dfrac{7}{3} x - 2$ The y-intercept for the second equation is $-2$ , so the second line must pass through the point $(0, -2)$ The slope for the second equation is $\dfrac{7}{3}$ . Remember that the slope tells you rise over run. So in this case for every $7$ positions you move up $3$ positions to the right. $7$ positions up from $(0, -2)$ is $(3, 5)$ Graph the green line so it passes through $(0, -2)$ and $(3, 5)$ The solution is the point where the two lines intersect. The lines intersect at $(3, 5)$.